can't set variable in JavaScript based on json id from php

Question

I have a simple array from a php file (the reason for not using a json file is cause this info comes from a mysql database)

$array[0] = array('id' => 1,'price' => '325');
$array[1] = array('id' => 2,'price' => '486');      
header('Content-type: application/json');
echo json_encode($array);

and it echos as:

[{"id":1,"price":"325"},{"id":2,"price":"486"}]

now what I want to do is take the id and add it to a variable called counterval so that JavaScript will read it as

counterval1 = 325;
counterval2 = 486;

but I can't seem to get it to read that way. Here is the script at the current moment.

$.getJSON('test.php',function(data) {
    $.each(data, function(i) {
    counterval + data[i].id = data[i].price;
    console.log (counterval2);
});
    $('#results').html(counterval2);    
});
var counterval1 = 0;
var counterval2 = 0;

any help on this would be greatly appreciated.

Why don't you use an array? `counterval[data[i].id]`? – Imperative Mar 18 '13 at 07:08 — Imperative, Mar 18 '13 at 07:08
Please update your title and tags to JavaScript – Michel Feldheim Mar 18 '13 at 07:09 — Michel Feldheim, Mar 18 '13 at 07:09

Reigel Gallarde · Accepted Answer · 2014-06-10T01:04:16.627

3

you can't do that... but you can do this...

var counterval = [];

$.getJSON('test.php',function(data) {
    $.each(data, function(i) {
         counterval[data[i].id] = data[i].price;
         console.log (counterval[2]);
    });
    $('#results').html(counterval[2]);    
});
counterval[1] = 0;
counterval[2] = 0;

edited Jun 10 '14 at 01:04

answered Mar 18 '13 at 07:05

Reigel Gallarde

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hmm... console.log seems to say an error of 'unexpected token' but I see what your getting at. – Keleko Mar 18 '13 at 07:10
maybe you don't have a `2` in there? – Reigel Gallarde Mar 18 '13 at 07:12
no, kept giving me the same error, I think it was having problems with the var counterval[1] = 0; – Keleko Mar 18 '13 at 07:17
ahh... how about changing `var counterval = [];` to `var counterval = {};` – Reigel Gallarde Mar 18 '13 at 07:25
hmm.. nope same output. I even tried copy pasting from your script. – Keleko Mar 18 '13 at 07:28
fixed it, it was as I thought. I removed the var at the bottem and the script executed as It should. Thank you so much for all your help – Keleko Mar 18 '13 at 07:31

score 2 · Answer 2 · answered Mar 18 '13 at 07:11

2

See my comment on your post. If you really want the vars to look like you explained:

eval("counterval" + data[i].id + " = data[i]..price");
alert(counterval1);

answered Mar 18 '13 at 07:11

Imperative

3,138
2
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This is correct, but it does what he wants. I wouldn't recommend it though. – Imperative Mar 18 '13 at 07:12
This does work very well with what I'm wanting, I would like to know why Imperative you say you would not recommend it? – Keleko Mar 18 '13 at 07:17
1

Read through this [SO thread](http://stackoverflow.com/questions/86513/why-is-using-the-javascript-eval-function-a-bad-idea) – Imperative Mar 18 '13 at 07:22
ah I see, what might be a safer route? rather then using eval. – Keleko Mar 18 '13 at 07:26

can't set variable in JavaScript based on json id from php

2 Answers2